Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 51
To see that no set with five or more vectors can be independent, set up
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
a 1,1 a 1,2 a 1,3 a 1,4 a 1,5 0
c 1
⎜a 2,1
⎟
⎝a 3,1
⎠ + c ⎜a 2,2
⎟
2 ⎝a 3,2
⎠ + c ⎜a 2,3
⎟
3 ⎝a 3,3
⎠ + c ⎜a 2,4
⎟
4 ⎝a 3,4
⎠ + c ⎜a 2,5
⎟
5 ⎝a 3,5
⎠ = ⎜0
⎟
⎝0⎠
a 4,1 a 4,2 a 4,3 a 4,4 a 4,5 0
and note that the resulting linear system
a 1,1 c 1 + a 1,2 c 2 + a 1,3 c 3 + a 1,4 c 4 + a 1,5 c 5 = 0
a 2,1 c 1 + a 2,2 c 2 + a 2,3 c 3 + a 2,4 c 4 + a 2,5 c 5 = 0
a 3,1 c 1 + a 3,2 c 2 + a 3,3 c 3 + a 3,4 c 4 + a 3,5 c 5 = 0
a 4,1 c 1 + a 4,2 c 2 + a 4,3 c 3 + a 4,4 c 4 + a 4,5 c 5 = 0
has four equations and five unknowns, so Gauss’ method must end with at least one c variable free,
so there are infinitely many solutions, and so the above linear relationship among the four-tall vectors
has more solutions than just the trivial solution.
The smallest linearly independent set is the empty set.
The biggest linearly dependent set is R 4 . The smallest is {⃗0}.
Two.II.1.36 (a) The intersection of two linearly independent sets S ∩T must be linearly independent
as it is a subset of the linearly independent set S (as well as the linearly independent set T also, of
course).
(b) The complement of a linearly independent set is linearly dependent as it contains the zero vector.
(c) We must produce an example. One, in R 2 , is
( (
1 2
S = { } and T = { }
0)
0)
since the linear dependence of S 1 ∪ S 2 is easily seen.
(d) The union of two linearly independent sets S ∪ T is linearly independent if and only if their
spans have a trivial intersection [S] ∩ [T ] = {⃗0}. To prove that, assume that S and T are linearly
independent subsets of some vector space.
For the ‘only if’ direction, assume that the intersection of the spans is trivial [S] ∩ [T ] = {⃗0}.
Consider the set S ∪ T . Any linear relationship c 1 ⃗s 1 + · · · + c n ⃗s n + d 1
⃗t 1 + · · · + d m
⃗t m = ⃗0 gives
c 1 ⃗s 1 + · · · + c n ⃗s n = −d 1
⃗t 1 − · · · − d m
⃗t m . The left side of that equation sums to a vector in [S], and
the right side is a vector in [T ]. Therefore, since the intersection of the spans is trivial, both sides
equal the zero vector. Because S is linearly independent, all of the c’s are zero. Because T is linearly
independent, all of the d’s are zero. Thus, the original linear relationship among members of S ∪ T
only holds if all of the coefficients are zero. That shows that S ∪ T is linearly independent.
For the ‘if’ half we can make the same argument in reverse. If the union S ∪ T is linearly
independent, that is, if the only solution to c 1 ⃗s 1 + · · · + c n ⃗s n + d 1
⃗t 1 + · · · + d m
⃗t m = ⃗0 is the trivial
solution c 1 = 0, . . . , d m = 0, then any vector ⃗v in the intersection of the spans ⃗v = c 1 ⃗s 1 +· · ·+c n ⃗s n =
−d 1
⃗t 1 − · · · = d m
⃗t m must be the zero vector because each scalar is zero.
Two.II.1.37 (a) We do induction on the number of vectors in the finite set S.
The base case is that S has no elements. In this case S is linearly independent and there is
nothing to check — a subset of S that has the same span as S is S itself.
For the inductive step assume that the theorem is true for all sets of size n = 0, n = 1, . . . , n = k
in order to prove that it holds when S has n = k+1 elements. If the k+1-element set S = {⃗s 0 , . . . , ⃗s k }
is linearly independent then the theorem is trivial, so assume that it is dependent. By Corollary 1.17
there is an ⃗s i that is a linear combination of other vectors in S. Define S 1 = S − {⃗s i } and note that
S 1 has the same span as S by Lemma 1.1. The set S 1 has k elements and so the inductive hypothesis
applies to give that it has a linearly independent subset with the same span. That subset of S 1 is
the desired subset of S.
(b) Here is a sketch of the argument. The induction argument details have been left out.
If the finite set S is empty then there is nothing to prove. If S = {⃗0} then the empty subset will
do.
Otherwise, take some nonzero vector ⃗s 1 ∈ S and define S 1 = {⃗s 1 }. If [S 1 ] = [S] then this proof
is finished by noting that S 1 is linearly independent.
If not, then there is a nonzero vector ⃗s 2 ∈ S − [S 1 ] (if every ⃗s ∈ S is in [S 1 ] then [S 1 ] = [S]).
Define S 2 = S 1 ∪ {⃗s 2 }. If [S 2 ] = [S] then this proof is finished by using Theorem 1.17 to show that
S 2 is linearly independent.